Question

1. An exponential relationship can be defined by the formula f(x)=30(0.42)^x where y=f(x). Complete each of the following exercises.

1. As x increases, y    answer:decreases

2. What is the function's initial value (the value of y when x=0)?

   f(0)=

3. What is the function's 1-unit growth or decay factor?

4. What is the function's 1-unit percent change?

5. What is the value of f(3.25)?

2.    The exponential function f(t)=18.6(0.964)^t where w=f(t) models the water level (in meters) in a certain reservoir in terms of the number of weeks since June 1.

1. As t increases, w    answer:decreases

2. What was the water level in the reservoir on June 1?

3. What is the function's 1-week growth or decay factor?

4. What is the function's 1-week percent change?

5. What was the water level in the reservoir 5.5 weeks after June 1

3.

1. In a certain exponential relationship, whenever x changes by 1, y changes by -4%. For this relationship, the 1-unit growth (decay) factor is

2. In a different exponential relationship, whenever x changes by 1, y changes by -22%. For this relationship, the 1-unit growth (decay) factor is

3. In a third exponential relationship, whenever x changes by 1, y changes by -57%. For this relationship, the 1-unit growth (decay) factor is

Answer 1

1.

a. In this case, the base is 0.42 < 1, this implies as x increases y decreases

b. When x = 0

=> f(0) = 30

c. 1-unit decay factor = 0.42

d. 1-unit percent change

= (1 - 0.42) * 100%

= 0.58 * 100 %

= 58 %

e. when x = 3.25

  

=> f(3.25) = 1.789293

2.

a. In this case, the base is 0.964 < 1, this implies as x increases y decreases

b. When x = 0

=> f(0) = 18.6

Therefore, the water level in the reservoir on June 1 = 18.6 m

c. 1-week decay factor = 0.964

d. 1-week percent change

= (1 - 0.964) * 100%

= 0.036 * 100 %

= 3.6 %

e. when t = 5.5

  

=> f(5.5) = 15.20326

Therefore, the water level in the reservoir 5.5 weeks after June 1 = 15.20326 m

3. a) r = - 4 % = - 0.04

Therefore,

1-unit decay factor

= 1 - 0.04

= 0.96

b) r = - 22 % = - 0.22

Therefore,

1-unit decay factor

= 1 - 0.22

= 0.78

c) r = - 57 % = - 0.57

Therefore,

1-unit decay factor

= 1 - 0.57

= 0.43